The fractional atomic weight of natural Chlorine is . This is primarily due to the existence of two — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Let the fractional abundance of ${}^{37}Cl$ be $y$. Then the fractional abundance of ${}^{35}Cl$ is $(1-y)$. The average atomic weight is $35(1-y) + 37y = 35.5$. Solving this yields $35 - 35y + 37y = 35.5 \Rightarrow 2y = 0.5 \Rightarrow y = 0.25$. Thus, ${}^{37}Cl$ is 25% and ${}^{35}Cl$ is 75%. The ratio of ${}^{35}Cl$ to ${}^{37}Cl$ is $75 : 25$, which simplifies to $3 : 1$. Therefore, $x = 3$.